Question

State Coulomb’s law and express it in vector form.

Answer 1

Hint: The electrostatic force between the stationary charges can be calculated using the Coulomb’s law. The electrostatic force between the charges is given by,
\[\Rightarrow F=k\dfrac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}\]

Complete step-by-step answer:
The Coulomb’s law of electrostatic force states that the electrostatic force of attraction and repulsion are directly proportional to the product of the charges and inversely proportional to the square of the distance between the charges.

\[\begin{align}
  & \Rightarrow F\propto {{q}_{1}}\times {{q}_{2}} \\
 & \Rightarrow F\propto \dfrac{1}{{{r}^{2}}} \\
\end{align}\]

Where, \[{{q}_{1}},{{q}_{2}}\] are the charges and r is the distance between the charges
Thus, the electrostatic force is given by,
\[\begin{align}
  & \Rightarrow F\propto \dfrac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}} \\
 & \Rightarrow F=k\dfrac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}} \\
\end{align}\]

Where, k is the constant of proportionality

The vector form of the Coulomb’s law is given as:
\[\Rightarrow \vec{F}=k\dfrac{{{q}_{1}}{{q}_{2}}}{r_{12}^{2}}{{\hat{r}}_{12}}\]

Here, ${{r}_{12}}$ is the displacement from charge 1 to charge 2.

Note:
The magnitude of force between charge 1 and charge 2 is equal. But in vector form force on charge 1 is given as:
\[\Rightarrow {{\vec{F}}_{12}}=k\dfrac{{{q}_{1}}{{q}_{2}}}{r_{12}^{2}}{{\hat{r}}_{12}}\]
Force on charge 2 is given as:
\[\Rightarrow {{\vec{F}}_{21}}=k\dfrac{{{q}_{1}}{{q}_{2}}}{r_{21}^{2}}{{\hat{r}}_{21}}\]

\[\Rightarrow {{\vec{F}}_{12}}={{\vec{F}}_{21}}\]